Return1since the palindrome partitioning["aa","b"]could be produced using 1 cut.本文地址 算法1:在上一题的基础上,我们很容易想到的是在DFS时,求得树的最小深度即可(遍历时可以根据当前求得的深度进行剪枝),但是可能是递归层数太多,大数据时运行超时,也贴上代码: View Code 算法2:设f[i][j]是i为起点...
https://leetcode.com/problems/palindrome-partitioning-ii/ 题目: Given a string s, partition s such that every substring of the partition is a palindrome. Return the minimum cuts needed for a palindrome partitioning of s. For example, given s = “aab”, Return 1 since the palindrome partitio...
如此就可以获得所有的划分。 2Palindrome Partitioning II 问题来源:Palindrome Partitioning II 该问题是问题I的变种,现在不求所有的划分,而只求分组个数最小的划分。 2.1 深搜 该问题一开始的思路肯定是直接照搬问题I的方法,采用DFS去求解。如果我们按照这个思路去实现,提交的时候会发现算法超时,即使进行了一系列的调...
[Leetcode][python]Palindrome Partitioning/Palindrome Partitioning II/分割回文串/分割回文串II Palindrome Partitioning 题目大意 将一个字符串分割成若干个子字符串,使得子字符串都是回文字符串,要求列出所有的分割方案。 解题思路 DFS 代码 class Solution(object): def partition(self, s): """ :type s: str ...
Return 1 since the palindrome partitioning["aa","b"]could be produced using 1 cut. 给定字符串s,分区的每个子字符串都是回文。返回s的回文分区所需的最小切割数。 例如,给定s=“aab”,返回1,因为回文分区[“aa”,“b”]可以切割1次生成
My code: publicclassSolution{publicintminCut(Strings){if(s==null||s.length()==0){return0;}intn=s.length();int[]mcut=newint[n];boolean[][]pal=newboolean[n][n];initialize(pal,s);for(inti=1;i<n;i++){if(pal[0][i]){mcut[i]=0;}else{intmin=mcut[i-1]+1;for(intj=i-...
Leetcode 132 Palindrome Partioning II 题目描述 Given a string s, partition s such that every substring of the partition is a palindrome.Return the minimum cuts needed for a palindrome partitioning o…
Given a strings, partitionssuch that every substring of the partition is apalindrome. Returnall possible palindrome partitioning ofs. Example 1: Input: s = "aab" Output: [["a","a","b"],["aa","b"]] Example 2: Input: s = "a" ...
【LeetCode】1616.Split Two Strings to Make Palindrome wisdompeak 37 播放 · 0 弹幕 分割回文串II 动态规划 - Palindrome Partition Dynamic Programming alchemist_dong 188 播放 · 0 弹幕 python算法-7穷竭搜索-10Leetcode 131 Palindrome Partitioning 千寻Coder 18 播放 · 0 弹幕 小白刷题之Palindrome相...
(2)设dp[i]为分割s为回文串的最小次数,状态转移方程为: dp[i]=min(dp[j]+1),当isPal[j+1][i]为true时; 这里有个关键点,整个串也算是自身的一个子串,纠结半天,觉得至少需要分割一次,即子串长度必须为[1,n-1],想多了。 classSolution{public:constintinf=0x3f3f3f3f;intminCut(string s){int...